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# If t and c are two complex numbers such that $|t| \ne |c|,|t| = 1{\text{ }}and{\text{ }}z = \dfrac{{at + b}}{{t - c}},z = x + iy.$ locus of z is(where a,b are complex number.A) Line segmentB) Straight lineC) CircleD) None of these  Hint: Since $|t| = 1$ we will try to separate the factor of ‘t’ from the expression $z = \dfrac{{at + b}}{{t - c}}$.
After separating the term ‘t’ we will take the modulus function of both the side and will eliminate ‘t’ using$|t| = 1$.

Complete step by step solution:
Given data:
$a,\;b,\;c,\; \text{and}\:{\text{ t}}\:{\text{are }}\:{\text{complex }}\:{\text{numbers }} \\ |t| = 1 \\ |t| \ne |c| \\ z = \dfrac{{at + b}}{{t - c}},z = x + iy. \\$
Now solving the given equation, i.e.
$z = \dfrac{{at + b}}{{t - c}} \\ \Rightarrow z(t - c) = at + b \\ \Rightarrow zt - at = b + cz \\ \Rightarrow t(z - a) = b + cz \\ \Rightarrow t = \dfrac{{b + cz}}{{z - a}} \\$
Now, applying the modulus function on both sides of the equation, we obtain
$\left| t \right| = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\ \sin ce{\text{ }}\left| t \right| = 1 \\ 1 = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\$
${\text{Now have also condition on}}$ ${\text{'c'}}$that $\left| {\text{c}} \right| \ne {\text{1}}$
$1 = \left| {\dfrac{{c\left( {\dfrac{b}{c} + z} \right)}}{{z - a}}} \right| \\ we{\text{ }}know{\text{ }}that,{\text{ }}\left| {ab} \right| = \left| a \right|\left| b \right| \\ \Rightarrow 1 = \left| {\dfrac{{\dfrac{b}{c} + z}}{{z - a}}} \right|\left| c \right| \\ \Rightarrow \dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + \dfrac{b}{c}}}{{z - a}}} \right| \\$
Since b, c are complex numbers, then $\dfrac{b}{c}$ be also a complex number let ‘p’
$\dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + p}}{{z - a}}} \right|$
We know that,
$\left| {\dfrac{{z - \alpha }}{{z - \beta }}} \right| = A$
When $A \ne 1$, then the locus of z is a circle.
Here also $\left| c \right| \ne 1$ i.e.,
$\left| {\dfrac{{z + p}}{{z - a}}} \right| \ne 1$
Therefore the locus of z is a circle.

(C) The circle is the correct option.

Note: It is well known that $\left| z \right| = a,$ ‘a’ being a constant is an equation of a circle.
From the equation
$z = \dfrac{{at + b}}{{t - c}}$
Applying the modulus function on both sides of the equation, we obtain
$\left| z \right| = \left| {\dfrac{{at + b}}{{t - c}}} \right|$
We can say that the right-hand side will come to be a constant as a, b, c and t are some fixed complex number.
We can write it as
$\left| z \right| = f$, where f is constant, comes out to be an equation of a circle.
View Notes
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